If the `K_b` value in the hydrolysis reaction
`B^(+)+H_2O hArr BOH +H^+`
is `1.0 xx 10^(-6)`, then the hydrolysis contant of the salt would be

Options

A. `1 xx 10^(-6)`
B. `1 xx 10^(-7)`
C. `1 xx 10^(-8)`
D. `1xx 10^(-9)`

Answer

To find the hydrolysis constant (K_h) of the salt given the K_b value for the hydrolysis reaction \( B^+ + H_2O \rightleftharpoons BOH + H^+ \), we can use the relationship between the hydrolysis constant (K_h), the base dissociation constant (K_b), and the ion product of water (K_w).

### Step-by-Step Solution:

1. **Identify the given values**:
- The base dissociation constant \( K_b = 1.0 \times 10^{-6} \).
- The ion product of water \( K_w = 1.0 \times 10^{-14} \) at 25°C.

2. **Use the relationship between K_h, K_b, and K_w**:
The relationship is given by:
\[ K_h = \frac{K_w}{K_b} \]

3. **Substitute the known values into the equation**:
\[ K_h = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-6}} \]

4. **Perform the division**:
\[ K_h = 1.0 \times 10^{-14} \div 1.0 \times 10^{-6} = 1.0 \times 10^{-8} \]

5. **Final Result**:
The hydrolysis constant \( K_h \) of the salt is:
\[ K_h = 1.0 \times 10^{-8} \]

### Summary:
The hydrolysis constant of the salt is \( 1.0 \times 10^{-8} \).

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